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台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-1-
Chapter 5Two Degree Freedom Systems
5台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-2-
Learning Objectives
Formulate (用公式表示) the equations of motion of two-degree-of-
freedom systems
Identify the mass, damping, and stiffness matrices from theequations of motion
Compute the eigenvalues or natural frequencies of vibration and themodal (形態上的) vectors
Determine the free-vibration solution using the known initialconditions
Understand the concepts of coordinate coupling and principalcoordinates
Determine the forced-vibration solutions under harmonic forces
Understand the concepts of self-excitation and stability of thesystem
Use the Laplace transform approach for solution of two-DOF systems
Solve two-DOF free- and forced-vibration problems using MATLAB
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-3-
Chapter Outline
5.1 Introduction
5.2 Equations of Motion for Forced Vibration
5.3 Free Vibration Analysis of an Undamped System
5.4 Torsional System
5.5 Coordinate Coupling and Principal Coordinates
5.6 Forced-Vibration Analysis
5.7 Semidefinite Systems
5.8 Self-Excitation and Stability Analysis
5.9 Transfer-Function Approach
5.10 Solutions Using Laplace Transform
5.11 Solutions Using Frequency Transfer Functions
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-4-
5.1Introduction
5.1
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-5-
5.1 Introduction
• Two-degree-of-freedom systems are defined as systems that
require two independent coordinates to describe their motion,
as introduced in Fig. 1.12.
Fig. 1.12 Two-degree-of-freedom systems
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-6-Fig. 5.1 Lathe
X1(t) X2(t)
(t)
X (t)
隆起團塊的
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-7-
Fig. 5.2 Automobile
X1(t)X2(t)
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-8-
Fig. 5.3 Multistory building
subjected to an earthquake
土地
多層的
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-9-
Fig. 5.4 Packaging of an instrument
Fig. 5.4(a) Illustrates the packaging of an instrument of mass m.Assuming that the motion of the instrument is confined to the xy-plane, the system can be modeled as a mass m supported by springsin the x and y directions, as shown in Fig. 5.4(b). Thus the systemhas one point mass m and two degrees of freedom, because themass has two possible types of motion (translations along the x andy directions).
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-10-
5.1 Introduction
• The general rule for the computation of the number of degrees of freedom can be stated as follows:
• There are two equations of motion for a two-DOF system, one for each mass (more precisely, for each degree of freedom). They are generally in the form of coupled differential equations—that is, each equation involves all the coordinates.
• If a harmonic solution is assumed for each coordinate, the equations of motion lead to a frequency equation that gives two natural frequencies for the system. If we give suitable initial excitation, the system vibrates at one of these natural frequencies.
No. of degrees of freedom of the system
No. of masses in the system x No. of possible types of motion of each mass
=
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-11-
5.1 Introduction
• During free vibration at one of the natural frequencies, the amplitudes of the two-DOFs (coordinates) are related in a specific manner and the configuration is called a normal mode, principal mode, or natural mode of vibration. Thus a two-DOF system has two normal modes of vibration corresponding to the two natural frequencies.
• If we give an arbitrary initial excitation to the system, the resulting free vibration will be a superposition of the two normal modes of vibration. However, if the system vibrates under the action of an external harmonic force, the resulting forced harmonic vibration take place at the frequency of the applied force.
• Under harmonic excitation, resonance occurs (i.e., the amplitudes of the two coordinates will be maximum) when the forcing frequency is equal to one of the natural frequencies of the system.
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
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台灣師範大學機電科技學系
C. R. Yang, NTNU MT
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5.2Equations of Motion for Forced Vibration
5.2台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-14-
5.2 Equations of Motion for Forced Vibration
• Consider a viscously damped two-degree-of-freedom spring-mass
system, shown in the figure below
f f
Fig. 5.5 A two-DOF spring-mass-damper system
k2=c2=0 uncouple
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-15-
5.2 Equations of Motion for Forced Vibration
• The application of Newton’s second law of motion to each of the masses gives the equations of motion:
1
1 2 2 1 2 2 1 1 1 1 1 1 1
2
2 2 2 1 2 2 1 3 2 3 2 2 2
:
( ) ( )
:
( ) ( )
F ma
For mass m
f k x x c x x k x c x m x
For mass m
f k x x c x x k x c x m x
1 1 1 2 1 2 2 1 2 1 2 2 1
2 2 2 1 2 3 2 2 1 2 3 2 2
( ) ( ) (5.1)
( ) ( ) (5.2)
m x c c x c x k k x k x f
m x c x c c x k x k k x f
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-16-
5.2 Equations of Motion for Forced Vibration
• To rewrite the equations of motion as:
• It can be seen that Eq. (5.1) contains terms involving x2
(i.e., ), whereas Eq. (5.2) contains terms involving x1
(i.e., ). Hence they represent a system of two coupled
second-order differential equations. We can therefore expect that the motion of the mass m1 will influence the motion of the mass m2, and vice versa.
1 1 1 2 1 2 2 1 2 1 2 2 1
2 2 2 1 2 3 2 2 1 2 3 2 2
( ) ( ) (5.1)
( ) ( ) (5.2)
m x c c x c x k k x k x f
m x c x c c x k x k k x f
2 2 2 2 c x and k x
2 1 2 1 c x and k x
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-17-
• Both equations (5.1 and 5.2) can be written in matrix form as
where [m], [c], and [k] are called the mass, damping, and stiffness matrices, respectively, and are given by
• It can be seen that the matrices [m], [c], and [k] are symmetric:
where the superscript T denotes the transpose of the matrix.
• And the displacement and force vectors are given respectively:
[ ] ( ) [ ] ( ) [ ] ( ) ( ) (5.3)m x t c x t k x t f t
1 2 2 1 2 21
2 2 3 2 2 32
0[ ] [ ] [ ]
0
c c c k k kmm c k
c c c k k km
1 1
2 2
( ) ( )( ) , ( )
( ) ( )
x t f tx t f t
x t f t
][][],[][],[][ kkccmm TTT
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-18-
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-19-
5.3Free-Vibration Analysis of an Undamped System
5.3台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-20-
5.3 Free-Vibration Analysis of an Undamped System
• The solution of Eqs.(5.1) and (5.2) involves four constants of
integration (two for each equation). We shall first consider the free
vibration solution of Eqs.(5.1) and (5.2).
• By setting f1(t) = f2(t) = 0, and damping disregarded, i.e., c1 = c2 =
c3=0, and the equation of motion is reduced to:
• Assuming that it is possible to have harmonic motion of m1 and m2
at the same frequency ω and the same phase angle Φ, but with
different amplitude, we take the solutions as
)5.5(0)()()()(
)4.5(0)()()()(
2321222
2212111
txkktxktxm
txktxkktxm
)6.5()cos()(
)cos()(
22
11
tXtx
tXtx
Eq. (5.1)
Eq. (5.2) f1(t)=f2(t)= 0
c1= c2=c3=0
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-21-
5.3 Free-Vibration Analysis of an Undamped System
• Substituting Eq. (5.6) into Eqs.(5.4) and (5.5), we obtain
• Since Eq.(5.7)must be satisfied for all values of the time t, the
terms between brackets must be zero. Thus,
which represent two simultaneous homogenous algebraic equations
in the unknown X1 and X2.
2
1 1 2 1 2 2
2
2 1 2 2 3 2
( ) cos( ) 0
( ) cos( ) 0 (5.7)
m k k X k X t
k X m k k X t
2
2 2 1 1 2
1 1 2 1 2 2
1 2
2 2 2
2 1 2 2 3 2 2
1 2 2 3
( )( ) 0
( ) 0 (5.8) ( )
X m k km k k X k X
X k
X kk X m k k X
X m k k
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-22-
5.3 Free-Vibration Analysis of an Undamped System
• For trivial solution, i.e., X1 = X2 = 0, which implies that there is no
vibration. For a nontrivial solution, the determinant of the
coefficients of X1 and X2 must be zero:
or
Eq. (5.9) is called the frequency or characteristic equation, because
its solution yields the frequencies or the characteristic values of the
system.
0)(
)(det
21
2
12
221
2
1
kkmk
kkkm
)9.5(0))((
)()()(
2
23221
132221
4
21
kkkkk
mkkmkkmm
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-23-
5.3 Free-Vibration Analysis of an Undamped System
• The roots of Eq. (5.9) are given by:
• The roots 1 and 2 are called natural frequencies of the system.
2 2 1 2 2 2 3 1
1 2
1 2
2
1 2 2 2 3 1
1 2
2
1/21 2 2 3 2
1 2
( ) ( )1,
2
( ) ( )1[
2
( )( )4 ] (5.10)
k k m k k m
m m
k k m k k m
m m
k k k k k
m m
2 21 21 2
1 2
, k k
m m
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-24-
5.3 Free-Vibration Analysis of an Undamped System
• To determine the values of X1 and X2, these values depend on the natural
frequencies 1 and 2. We shall denote the values of X1 and X2 corresponding
to 1 as X1(1) and X2
(1) and those corresponding to 2 as X1(2) and X2
(2). From
Eq. (5.8) gives
• The normal modes of vibration corresponding to ω12 and ω2
2 can be
expressed, respectively, as
• The vectors , which denote the normal modes of vibration, are
known as the modal vectors of the system.
2 1 1 2 2
1
1 2 2 2 3
2 1 1 2 2
2
1 2 2 2 3
(2) 2
(1) 2
1
(1) 2
2
2
1
(2)
2
( )
( )
( )(5.11)
( )
X m k k kr
X k m k k
X m k k kr
X k m k k
)12.5( and )2(
12
)2(
1
)2(
2
)2(
1)2(
)1(
11
)1(
1
)1(
2
)1(
1)1(
Xr
X
X
XX
Xr
X
X
XX
(1) (2) X and X
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-25-
5.3 Free-Vibration Analysis of an Undamped System
• The free-vibration solution or the motion in time can be expressed,
using Eq. (5.6), as
Where the constants are determined by the initial
conditions.
(1) (1)
1 1 1(1)
(1) (1)
2 1 1 1
(2) (2)
1 1 2(2)
(2) (2)
2 2 1
1
2 2
1
2
( ) cos( )( )
( ) cos( )
( ) cos( )( ) (5.13)
(second m
first mo
o) cos( )
de
de
x t X tx t
x t r X t
x t X tx t
x t r X t
(1) (2)
1 1 1 2, , , X X and
1 1 1
2 2 2
( ) cos( ) for
( ) cos( ) for (5.6)
x t X t m
x t X t m
(1)
(1)
(2
2
1
1
2
2
)
(2)
1
(5.11)
Xr
X
Xr
X
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-26-
• Initial conditions
As stated in Section 5.1, the system can be made to vibrate in its i-th
normal mode (i=1, 2) by subjecting it to the specific initial conditions.
However, for any other general initial conditions, both modes will be
excited. The resulting motion, which is given by the general solution of Eqs.
(5.4) and (5.5), can be obtained by a linear superposition of the two
normal modes, Eq.(5.13):
where c1 and c2 are constants. Since already involve the
unknown constants (see Eq. (5.13)), we can choose c1=c2=1
with no loss of generality.
1 1 1
2 1 2
( )
( )
( 0) some constant, ( 0) 0,
( 0) , ( 0) 0
i
i
i
x t X x t
x t r X x t
1 1 2 2( ) ( ) ( ) (5.14)x t c x t c x t
(1) (2)( ) ( )x t and x t
(1) (2)
1 1 X and X
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-27-
5.3 Free-Vibration Analysis of an Undamped System
• Thus, the components of the vector can be expressed, using
Eq. (5.14) with c1=c2=1 and Eq. (5.13), as
• The unknown constants can be determined from
the initial conditions:
(1) (2)
2 2
(1) (2)
1
(1) (2) (1) (2)
1 1 1 1 1 1 1 2
1 2 1
2
(1) (2)
2 2 2 1 1 2 2
1 1 2 2
( ) ( ) ( ) cos( ) cos( )
( ) ( ) ( ) cos( ) cos( )
cos( ) cos( ) (5.15)
X X
r X r
x t x t x t X t X t
x t x t x t t t
t tX
1 1 1 1
2 2 2 2
( 0) (0), ( 0) (0),
( 0) (0), ( 0) (0) (5.16)
x t x x t x
x t x x t x
( )x t
(1) (2)
1 1 1 2, , , X X and
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-28-
5.3 Free-Vibration Analysis of an Undamped System
• Substituting Eq. (5.16) into Eq.(5.15) leads to
• The solution of Eq. (5.17) can be expressed as
)17.5(sinsin)0(
coscos)0(
sinsin)0(
coscos)0(
2
)2(
1221
)1(
1112
2
)2(
121
)1(
112
2
)2(
121
)1(
111
2
)2(
11
)1(
11
XrXrx
XrXrx
XXx
XXx
)(
)0()0(sin,
)(
)0()0(sin
)0()0(cos,
)0()0(cos
122
2112
)2(
1
121
2121
)1(
1
12
2112
)2(
1
12
2121
)1(
1
rr
xxrX
rr
xxrX
rr
xxrX
rr
xxrX
Four algebraic equations
in the unknowns
(1)
1 1
(2)
1 2
(1)
1 1
(2)
1 2
cos
cos
sin
sin
X
X
X
X
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-29-
5.3 Free-Vibration Analysis of an Undamped System
• We can obtain the desired solution as
1/22 2
(1) (1) (1)
1 1 1 1 1
1/22 2
(2) (2) (2)
1 1
1/22
2 2 1 2
2 1 2 2
2 1 1
1/22
2 1 1 2
1 1
2 1 2
(1)
1 1
1
2 2
2 1 2
(0) (0)1(0) (0)
( )
(0) (0
cos sin
cos si
)1(0)
n
sta
)
n
(0)(
X X X
X X
r x xr x x
r r
r x xr x x
r r
X
X
1
(1)
1 1
(2)
1 1 2
2 (2)
1
1 2 1 2
1 2 1 2
1 1 1 2
2 1 1 22
in
cos
sinta
(0) (0)tan
[ (0) (0)
(0) (0)tan (5.18)
[ (0) (0n
c s )o
r x x
r x x
r x x
r x x
X
X
X
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-30-
Example 5.1 Frequencies of Spring-Mass System
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-31-
Fig. 5.6 Two-DOF system
2 21 21 2
1 2
, k k
m m
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-32-
(1) (1)
2 1 1X r X
(2) (2)
2 2 1X r X
for m1 vibration
for m2 vibration
for m1 vibration
for m2 vibration
under 1
under 2
superposition
superposition
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-33-
for m1 vibration
for m2 vibration
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-34-
5.3 Free-Vibration Analysis of an Undamped System
Example 5.3
Free-Vibration Response of a Two-Degree-of-Freedom System
Find the free-vibration response of the system shown in Fig.5.5(a)
with k1 = 30, k2 = 5, k3 = 0, m1 = 10, m2 = 1 and c1 = c2 = c3 = 0 for
the initial conditions .1 1 2 2(0) 1, (0) (0) (0) 0x x x x
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-35-
5.3 Free-Vibration Analysis of an Undamped System
Example 5.3
Free-Vibration Response of a Two Degree of Freedom System
Solution
For the given data, the eigenvalue problem, Eq.(5.8), becomes
By setting the determinant of the coefficient matrix in Eq.(E.1) to zero,
we obtain the frequency equation,
(E.1)0
0
55-
5 3510
0
0
2
1
2
2
2
1
32
2
22
221
2
1
X
X
X
X
kkmk
kkkm
4 210 85 150 0 (E.2)
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-36-
5.3 Free-Vibration Analysis of an Undamped System
Example 5.3
Free-Vibration Response of a Two-Degree-of-Freedom System
Solution
The natural frequencies can be found as
The substitution of in Eq. (E.1) leads to ,
while yields . Thus the normal modes
(or eigenvectors) are given by
2 2
1 2 1 22 5 6 0 1 5811 2 4495 or E.3). , . . , . (
1 2
1 11 1 2 2
1 11 2
2 2
1 1
2 5
X X X X X X E.4) E.
X5
X)
( ) ( )
( ) ( ) ( ) ( )
( ) ( )( (
2 2
1 2 5. 1 1
2 12X X( ) ( )
2 2
2 6 0. 2 2
2 15X X( ) ( )
(或1與2代入Eq. (5.11)求得r1與r2)
(1與2代入 (E.1)求得X1=rX2)
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-37-
5.3 Free-Vibration Analysis of an Undamped System
Example 5.3
Free-Vibration Response of a Two-Degree-of-Freedom System
Solution
The free-vibration responses of the masses m1 and m2 are given by
(see Eq.5.15):
By using the given initial conditions in Eqs.(E.6) and (E.7), we obtain
(E.7))4495.2cos(5)5811.1cos(2)(
(E.6))4495.2cos()5811.1cos()(
2
)2(
11
)1(
12
2
)2(
11
)1(
11
tXtXtx
tXtXtx
(1) (2)
1 1 1 1 2
(1) (2)
2 1 1 1 2
(1) (2)
1 1 1 1 2
(1) (2)
2 1 1 1 2
( 0) 1 cos cos (E.8)
( 0) 0 2 cos 5 cos (E.9)
( 0) 0 1.5811 sin 2.4495 sin (E.10)
( 0) 0 3.1622 sin 12.2475 sin (E.11)
x t X X
x t X X
x t X X
x t X X
The unknown constants
can be determined from
the initial conditions:
X X and(1) (2)
1 1 1 2, , ,
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-38-
5.3 Free-Vibration Analysis of an Undamped System
Example 5.3
Free-Vibration Response of a Two-Degree-of-Freedom System
Solution
The solution of Eqs.(E.8) and (E.9) yields
The solution of Eqs.(E.10) and (E.11) leads to
Equations (E.12) and (E.13) gives
(E.12)7
2cos;
7
5cos 2
)2(
11
)1(
1 XX
(1) (2)
1 1 1 2 1 2sin 0, sin 0 sin 0, sin 0 (E.13) X X
(E.14)0,0,7
2,
7
521
)2(
1
)1(
1 XX
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-39-
5.3 Free-Vibration Analysis of an Undamped System
Example 5.3
Free-Vibration Response of a Two-Degree-of-Freedom System
Solution
Thus the free vibration responses of m1 and m2 are given by
(E.16)4495.2cos7
105811.1cos
7
10)(
(E.15)4495.2cos7
25811.1cos
7
5)(
2
1
tttx
tttx
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-40-
5.4Torsional System
5.4
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-41-
5.4 Torsional System
• Consider a torsional system as shown in Fig.5.8. The three segments of the
shaft have rotational spring constants kt1, kt2, and kt3, The mass moments of
inertia of discs are J1 and J2, the applied torques Mt1 and Mt2, and the
rotational degrees of freedom 1 and 2.
• The differential equations of rotational motion for the discs can be derived as
1 1 1 1 2 2 1 1
2 2 2 2 1 3 2 2
( )
( )
t t t
t t t
J k k M
J k k M
Fig. 5.8 Torsional system with discs mounted on a shaft
M J
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-42-
• Upon rearrangement become
• For the free vibration analysis of the
system, Eq.(5.19) reduces to
• Note that Eq. (5.20) is similar to Eqs. (5.4) and (5.5). In fact, Eq.
(5.20) can be obtained by substituting 1, 2, J1, J2, kt1, kt2, and kt3
for x1, x2, m1, m2, k1, k2, and k3 , respectively. Thus the analysis
presented in Section 5.3 is also applicable to torsional systems.
)20.5(0)(
0)(
2321222
2212111
ttt
ttt
kkkJ
kkkJ
)19.5()(
)(
22321222
12212111
tttt
tttt
MkkkJ
MkkkJ
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-43-
5.4 Torsional System
Example 5.4
Natural Frequencies of a Torsional System
Find the natural frequencies and mode shapes for the torsional system
shown in the figure below for J1 = J0 , J2 = 2J0 and kt1 = kt2 = kt .
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-44-
5.4 Torsional System
Example 5.4
Natural Frequencies of a Torsional System
Solution
The differential equations of motion, Eq.(5.20), reduce to (with kt3 = 0,
kt1 = kt2 = kt, J1 = J0 and J2 = 2J0):
Rearranging and substituting the harmonic solution:
(E.1) 02
02
2120
2110
tt
tt
kkJ
kkJ
( ) cos( ); 1,2 (E.2)i i
t t i
1 1 1 2 1 2 2
2 2 2 1 2 3 2
( ) 0
( ) 0 (5.20)
t t t
t t t
J k k k
J k k k
Matrix
form
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-45-
5.4 Torsional System
Example 5.4
Natural Frequencies of a Torsional System
Solution
This gives the frequency equation of
The solution of Eq.(E.3) gives the natural frequencies
4 2 2 2 2 2
0 0 1 2
0 0
(5 17) (5 17)2 5 0 (E.3) ,
4 4
t t
t t
k kJ J k k
J J
1 2
0 0
(5 17) and (5 17) (E.4)4 4
t tk k
J J
台灣師範大學機電科技學系
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-46-
5.4 Torsional System
Example 5.4
Natural Frequencies of a Torsional System
Solution
The amplitude ratios are given by
Equations (E.4) and (E.5) can also be obtained by substituting the
following values into Eqs.(5.10) and (5.11).
(E.5)4
)175(2
4
)175(2
)2(
1
)2(
22
)1(
1
)1(
21
r
r
0and2,
,,
3022011
2211
kJJmJJm
kkkkkk tttt
Similar to Eq. (5.13), we can obtain-First mode and Second mode
台灣師範大學機電科技學系
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-47-
Similar to Eq. (5.13), we can obtain-First mode and Second mode
Example 5.4
Natural Frequencies of a Torsional System
Solution
Where the constants are determined by the initial
conditions.
(1) (1)
1 1 1(1)
(1) (1)
2 1 1 1
(2) (2)
1 1 2(2)
(2) (2)
2 2 1
1
2 2
1
2
( ) cos( )( )
( ) cos( )
( ) cos( )( ) (5.13)
( ) cos(second mode
first od
)
m et t
tt r t
t tt
t r t
(1) (2)
1 1 1 2, , , and
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-48-
1 2
0 0
5.4
(5 17) and (5 17) (E.4)4 4
t t
Example
k k
J J
台灣師範大學機電科技學系
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-49-
5.5Coordinate Coupling and Principal Coordinates
5.5台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-50-
5.5 Coordinate Coupling and Principal Coordinates
(一般化座標)
尾座
頭座
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-51-
Fig. 5.11 Lathe
尾座頭座
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-52-
Fig. 5.12 Modeling of a lathe
Generalized coordinates
有四種表示方式:
1 2
1
( ), ( ); ( ), ( )
( ), ( ); ( ), ( )
x t x t x t t
x t t y t t
台灣師範大學機電科技學系
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-53-
5.5 Coordinate Coupling and Principal Coordinates
Equations of Motion using x(t) and (t) (支點在C.G.):
From the free-body diagram shown in Fig. 5.12 (a), with the
positive values of the motion variables as indicated, the force
equilibrium equation in the vertical direction can be written as
And the moment equation about C.G. can be expressed as
Eqs.(5.21) and (5.22) can be rearranged and written in matrix form
as
1 1 2 2 1 2 2 2 1 1(5.2( ) ( ) ( ) 01) ( )mx k x l k x l mx k l k lk k x
2 2
2 2
0 1 1 1 2 2 2 0 1 11 1 2 2(5( ) ( ) ( ).22) 0( )J k x l l k x l k l kl J k l k ll x
)23.5(0
0
)( )(
)( )(
0
0 2
2
2
12211
221121
0 21
x
lklklklk
lklkkkx
J
m
(向下為正)
(順時針為正)
台灣師範大學機電科技學系
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-54-
Equations of motion Using y(t) and θ(t) (支點在P):
From Fig. 5.12(b), where y(t) and θ(t) are used as the generalized
coordinates of the system, the equations of motion for translation and
rotation can be written as
1 1 2 2 1 2 2 2 1 1 0my k y l k y l me my me k k y k l k l( ) ( ) ( ) ( )
2 2
1 1 1 2 2 2 1 1 2 2 1 1 2 2 0P PJ k y l l k y l l mey J mey k l k l y k l k l( ) ( ) ( ) ( )
(向下為正)
(順時針為正)
作用在C.G.點的轉矩
作用在C.G.點的力
台灣師範大學機電科技學系
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-55-
5.5 Coordinate Coupling and Principal Coordinates
These equations can be rearranged and written in matrix form as
)25.5(0
0
)()(
)()(
2
2
2
112211
112221
2
y
lklklklk
lklkkky
Jme
mem
P
me mey即有 和 不等於零項
=mass (m) × tangential acceleration ( )
=force ( ) × moment arm (e)my
e
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-56-
固有的
台灣師範大學機電科技學系
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-57-
5.5 Coordinate Coupling and Principal Coordinates
Example 5.6
Principal Coordinates of Spring-
Mass System
Determine the principal coordinates
for the spring-mass system shown in
the figure 5.6.
Fig. 5.6 Two-DOF system
台灣師範大學機電科技學系
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-58-
5.5 Coordinate Coupling and Principal Coordinates
Example 5.6 Principal Coordinates of Spring-Mass SystemSolution
Approach: Define two independent solutions as principal coordinates and express them in terms of the solutions x1(t) and x2(t).
The general motion of the system shown in Fig. 5.6 is given by Eq. (E.10) of Example 5.1:
Where are constants.
(E.1)3
coscos)(
3coscos)(
22112
22111
tm
kBt
m
kBtx
tm
kBt
m
kBtx
(1) (2)
1 1 2 1 1 2, , , B X B X and
台灣師範大學機電科技學系
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-59-
5.5 Coordinate Coupling and Principal Coordinates
Example 5.6 Principal Coordinates of Spring-Mass System
Solution
We define a new set of coordinates q1(t) and q2(t) such that
Since q1(t) and q2(t) are harmonic functions, their corresponding
equations of motion can be written as
1 1 1
2 2 2
( ) cos
3( ) cos (E.2)
kq t B t
m
kq t B t
m
1 1
2 2
0
30 (E.3)
kq q
m
kq q
m
These equations represent a two-DOF system
whose natural frequencies are
.
Because there is neither static nor dynamic
coupling in the equations of motion (E.3), q1(t) and q2(t) are principal coordinates.
1 2 3k m and k m
Note that the equation of motion corresponding to the solution q=Bcos(t+) is given by 2 0q q
台灣師範大學機電科技學系
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-60-
5.5 Coordinate Coupling and Principal Coordinates
Example 5.6 Principal Coordinates of Spring-Mass System
Solution
From Eqs.(E.1) and (E.2), we can write
The solution of Eqs.(E.4) gives the principal coordinates:
(E.4))()()(
)()()(
212
211
tqtqtx
tqtqtx
(E.5))]()([2
1)(
)]()([2
1)(
212
211
txtxtq
txtxtq
台灣師範大學機電科技學系
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5.6Forced-Vibration Analysis
5.6台灣師範大學機電科技學系
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-62-
5.6 Forced-Vibration Analysis
• The equations of motion of a general two-degree-of-freedom system under external forces can be written as
Eqs. (5.1) and (5.2) can be seen to be special cases of Eq. (5.27), with m11=m1, m22=m2, and m12=0.
• Consider the external forces to be harmonic:
where ω is the forcing frequency.
• We can write the steady-state solutions as
Where X1 and X2 are, in general, complex quantities that depend on and the system parameters.
)27.5(
2
1
2
1
2221
1211
2
1
2221
1211
2
1
2212
1211
F
F
x
x
kk
kk
x
x
cc
cc
x
x
mm
mm
0 1 2 5 28i t
j jF t F e j( ) , , ( . )
)29.5(2,1,)( jeXtx ti
jj
台灣師範大學機電科技學系
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-63-
• Substitution of Eqs. (5.28) and (5.29) into Eq. (5.27) leads to
as in Section 3.5 (Page 278), we define the mechanical impedance Zrs(i) as
2 2
11 11 11 12 12 12 1
2 2212 12 12 22 22 22
10
20
( ) ( )
( ) ( )
(5.30)
m i c k m i c k X
Xm i c k m i c k
F
F
rs rs rs rsZ i m i c k r s2
( ) , 1,2 (5.31)
5.6 Forced-Vibration Analysis
台灣師範大學機電科技學系
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-64-
5.6 Forced-Vibration Analysis
• We can write Eq.(5.30) as:
where
• Eq.(5.32) can be solved to obtain:
)32.5()( 0FXiZ
20
10
0
2
1
2212
1211matrix Impedance
)( )(
)( )()(
F
FF
X
XX
iZiZ
iZiZiZ
1
05 33
( ) ( . )X FZ i
台灣師範大學機電科技學系
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-65-
5.6 Forced-Vibration Analysis
• The inverse of the impedance matrix is given by
• Eqs.(5.33) and (5.34) lead to the solution
• By substituting Eq. (5.35) into Eq. (5.29), we can find the complete
solution, x1(t) and x2(t).
1 22 12
2
11 22 12 12 11
1 Z i - Z iZ i
Z i Z i Z i Z i Z i
( ) ( )( )
( ) ( ) ( ) ( ) ( )
22 10 12 201 2
11 22 12
12 10 11 202 2
11 22 12
( ) ( )( )
( ) ( ) ( )
( ) ( )( ) (5.35)
( ) ( ) ( )
Z i F Z i FX i
Z i Z i Z i
Z i F Z i FX i
Z i Z i Z i
22 121
21 11
1
det
a a
a aA
A
1
0 5 33( .( ) )X Z i F
( ) , 1,2 (5.29)i t
j jx t X e j
台灣師範大學機電科技學系
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-66-
5.6 Forced-Vibration Analysis
Example 5.8
Steady-State Response of Spring-
Mass System
Find the steady-state response of system
shown in Fig.5.15 when the mass m1 is
excited by the force F1(t) = F10 cos ωt.
Also, plot its frequency response curve.
Solution
The equations of motion of the system
can be expressed as (from Eq. (5.3))
Fig. 5.15 A two-mass system subjected to harmonic force.
1 1 10
2 2
cos 0 2 (E.1)
0 2 0
x x F tm k - k
m x -k k x
台灣師範大學機電科技學系
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-67-
5.6 Forced-Vibration Analysis
Example 5.8 Steady-State Response of Spring-Mass System
Solution
Comparison of Eq. (E.1) with Eq. (5.27) shows that
We assume the solution to be as follows
Eq.(5.31) gives
( ) cos ; 1,2 (E.2)j jx t X t j
2
11 22 12( ) ( ) 2 , ( ) (E.3)Z Z m k Z k
m m m m c c c
k k k k k F F t F
11 22 12 11 12 22
11 22 12 1 10 2
, 0, 0,
2 , , cos , 0
Since F10cost=Re(F10eit), we
shall assume the solution also to
be xj=Re(Xjeit)=Xjcost, j=1, 2.
台灣師範大學機電科技學系
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-68-
5.6 Forced-Vibration Analysis
Example 5.8 Steady-State Response of Spring-Mass System
Solution
Hence X1 and X2 are given by Eq. (5.35)
By defining , Eqs.(E.4) and (E.5) can be expressed
as (E.6) and (E.7).
2 2
10 101 2 2 2 2 2
10 102 2 2 2 2 2
( 2 ) ( 2 )( ) (E.4)
( 2 ) ( 3 )( )
( ) (E.5)( 2 ) ( 3 )( )
m k F m k FX
m k k m k m k
kF kFX
m k k m k m k
k m and k m2 2
1 2 3
台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-69-
E.7)(
1
)(2
1
2
1
2
1
2
102
k
FX
2
10
1
1 2 2 2
2
1 1 1
2
( ) (E.6)
1
F
X
k
台灣師範大學機電科技學系
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-70-
Fig. 5.16 Frequency-response curves of example 5.8
台灣師範大學機電科技學系
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-71-
5.7Semidefinite Systems
5.7台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-72-
5.7 Semidefinite (半定) Systems
退化未拘束
奇異的(行列式=0)
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-73-
5.7 Semidefinite Systems
Fig. 5.17 Semidefinite systems with two-DOF台灣師範大學機電科技學系
C. R. Yang, NTNU MT
-74-
5.7 Semidefinite Systems
• For Fig. 5.17 (a) and (b), the equations of motion can be written as
• For free vibration, we assume the motion to be harmonic:
• Substituting Eq.(5.37) into Eq.(5.36) gives
1 1 2 1 1 1 1 2
2 2 2 1 2 2 2 1
( ) ( ) 0
( ) ( ) 0 (5.36)
m x k x x m x k x x
m x k x x m x k x x
)37.5(2,1),cos()( jtXtx jjj
)38.5(0)(
0)(
2
2
21
21
2
1
XkmkX
kXXkm
台灣師範大學機電科技學系
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5.7 Semidefinite Systems
• By equating the determinant of the coefficients of X1 and X2 to zero,
we obtain the frequency equation as
• From which the natural frequencies can be obtained:
)39.5(0)]([ 21
2
21
2 mmkmm
)40.5()(
and 021
2121
mm
mmk
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-76-
Example 5.9 Free Vibration of an Understrained System
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-78-
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-79- 台灣師範大學機電科技學系
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-80-
5.8Self-Excitation and Stability Analysis
5.8
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-81-
5.8 Self-Excitation and Stability Analysis
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-82-
台灣師範大學機電科技學系
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-83-
四次方的
羅斯 - 赫維茲
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The same with Eq. (5.46)
(羅斯 -赫維茲準則)Routh-Hurwitz criterion
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(羅斯 -赫維茲準則)Routh-Hurwitz criterion
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0 2
1 3
1
a a
a aA
a
0 4
1 5
1
a a
a aB
a
0 6
1
6
1
0
a a
aa
a
1 3
a a
A BC
A
1 5
6
a a
A aD
A
1 0
00
a
A
A
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-88-
5.9Transfer-Function Approach
5.9
台灣師範大學機電科技學系
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5.9 Transfer-Function Approach
• For two-DOF system shown in Fig. 5.5, the equations of motion are
(Eqs. (5.1) and (5.2)):
• By taking the Laplace transforms of Eqs. (5.50) and (5.51),
assuming zero initial conditions, we obtain
5.51
5.50
2122321223222
1221212212111
fxkxkkxcxccxm
fxkxkkxcxccxm
5.53
5.52
212232122322
2
2
122121221211
2
1
sFsXksXkkssXcssXccsXsm
sFsXksXkkssXcssXccsXsm
台灣師範大學機電科技學系
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• Eqs. (5.52) and (5.53) can be rearranged to obtain
• Eqs. (5.54) and (5.55) indicate two simultaneous linear algebraic
equation in X1(s) and X2(s).These can be solved using Cramer’s rule
as
2
1 1 2 1 2 2 2 1
2
2 2
1
2
2
1 2 3 3 22 2
[ ] ( ) 5.54
( ) [ ) 5.53
m s c c s k k c s k F s
c s k m s c c s k
X s X s
X s sk sX F
5.9 Transfer-Function Approach
1 2
1
2
2
1 5.56 and
(t) (
5.57
)Apply inverse Laplace transfor
D s D sX s X
ms x and x
sD s D s
t
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5.9 Transfer-Function Approach
In Eqs. (5.56) and (5.57)
台灣師範大學機電科技學系
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5.9 Transfer-Function Approach
• Note that
1. The denominator, D(s), in the expressions of X1(s) and X2(s)
given by Eq. (5.60), is a fourth-order polynomial in s and
denotes the characteristic polynomial of the system. Because
the characteristic polynomial is of order four, the model (or
system) is said to be a fourth-order model (or system).
2. Equations (5.56) and (5.57) permit us to apply inverse Laplace
transforms to obtain the fourth-order differential equations for
x1(t) and x2(t) , respectively.
3. Equations (5.56) and (5.57) can be used to derive the transfer
functions of x1(t) and x2(t) corresponding to any specified
forcing function.
1 2
1 2 : and X s X s
Transfer function H s H sF s F s
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5.10Solutions Using Laplace Transform
5.10
The computation of response of two-degree-of-freedom systems using the Laplace transform.
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5.10Solutions Using Laplace Transform
振動問題
解析
建模聯立微分方程式
(ODEs)
(t的函數)
代數方程式(s的函數)
(僅需加減乘除運算,可簡化求解的程序)
求解1
X1(s)與X2(s)
求解2
x1(t)與x2(t)
解釋物理現象
Cramer’s rule
求解
取£
取£-1
部份分式展開轉換公式應用
£[x]
£[x]
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5.10 Solutions Using Laplace Transform
Example 5.12
Response Under Impulse Using Laplace Transform Method
Two railway cars, of masses m1 = M and m2 = m are connected by a
spring of stiffness k, as shown in the figure. If the car of mass M is
subjected to an impulse , determine the time responses of the
cars using the Laplace transform method.
tF 0
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5.10 Solutions Using Laplace Transform
Example 5.12
Response Under Impulse Using Laplace Transform Method
Solution
The responses of the cars can be determined using either of the
following approaches:
a. Consider the system to be undergoing free vibration due to the
initial velocity caused by the impulse applied to car M.
b. Consider the system to be undergoing forced vibration due to the
force applied to car M (with the displacements and velocities
of cars M and m considered to be zero initially). tF 0
台灣師範大學機電科技學系
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-98-
5.10 Solutions Using Laplace Transform
Example 5.12
Response Under Impulse Using Laplace Transform Method
Solution
Using the second approach, the equations of motion of the cars can be
expressed from Eq. (5.36)
Using the Laplace transforms, Eqs. (E.1) and (E.2) can be written as
1 1 2 0
2 2 1
E.1
0 E.2
Mx k x x F t
mx k x x
E.4 0
E.3
2
2
1
021
2
sXkmsskX
FskXsXkMs
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5.10 Solutions Using Laplace Transform
Example 5.12
Response Under Impulse Using Laplace Transform Method
Solution
Equations (E.3) and (E.4) can be solved by Cramer’s rule
E.6
E.5
22
02
22
2
01
mMkMmss
kFsX
mMkMmss
kmsFsX
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5.10 Solutions Using Laplace Transform
Example 5.12
Response Under Impulse Using Laplace Transform Method
Solution
Using partial fractions, Eqs. (E.5) and (E.6) can be rewritten as
where
E.8 11
E.7 1
222
02
222
01
ws
w
wsmM
FsX
ws
w
wM
m
smM
FsX
E.9 112
mMkw
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5.10 Solutions Using Laplace Transform
Example 5.12
Response Under Impulse Using Laplace Transform Method
Solution
The inverse transforms of Eqs. (E.7) and (E.8), using the results of
Appendix D, yield the time responses of the cars as
01
02
sin E.10
1sin E.11
F mx s t wt
M m wM
Fx s t wt
M m w
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5.11Solutions Using Frequency Transfer Functions
5.11
The frequency transfer function can be obtained by substituting i in place of s in the general transfer function of the system.
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5.11 Solutions Using Frequency Transfer Functions
Example 5.13
Derivation of Frequency
Transfer Functions
Derive the frequency transfer
functions of x1(t) and x2(t) for
the system shown in figure.
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5.11 Solutions Using Frequency Transfer Functions
Example 5.13
Derivation of Frequency Transfer Functions
Solution
From the free-body diagrams of the masses, the equations
of motion of the system is
E.2 0
E.1 sin
212212222
01212212111111
pxxkxxcxm
wtPpxxkxxcxkxcxm
(向下為正)
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5.11 Solutions Using Frequency Transfer Functions
Example 5.13
Derivation of Frequency Transfer Functions
Solution
By taking the Laplace transforms of Eqs. (E.1) and (E.2), assuming
zero initial conditions,
E.4 0
E.3
12212222
121221211111
2
1
sXsXksXsXcsXm
sPsXsXksXsXcsXkssXcsXsm
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5.11 Solutions Using Frequency Transfer Functions
Example 5.13
Derivation of Frequency Transfer Functions
Solution
The solutions X1(s) and X2(s) of Eqs. (E.3) and (E.4) are
where
E.6 and E.5 22
11
sD
sDsX
sD
sDsX
2
1 2 2 2
12 2
1
2
E.7
E.8
D s m s c s k
D s
P s
c s k P s
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5.11 Solutions Using Frequency Transfer Functions
Example 5.13
Derivation of Frequency Transfer Functions
Solution
We have
The general transfer functions is
E.9 211221
2
21221221
3
221221
4
22
kkskckc
scckmkmkmscmcmcmsmmsD
2
21 22
1 1
2 2 2 E.9 and E.10X s X s
P s P
m s c s k c s k
D s D ss
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C. R. Yang, NTNU MT
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5.11 Solutions Using Frequency Transfer Functions
Example 5.13
Derivation of Frequency Transfer Functions
Solution
The frequency transfer functions (s → i) of x1(t) and x2(t) is
where
E.13 and E.12 22
1
222
2
2
1
1
iwD
kiwc
iwP
iwX
iwD
kiwcwm
iwP
iwX
21122121221221
2
221221
4
21
4
kkkckciwcckmkmkmw
cmcmcmiwwmmwiwD
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5.12 Examples using MATLAB
•To practice by yourself from Ex. 5.15 to Ex.5.21
•The source codes of all MATLAB programs are given at the companion website
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